Probability of winning in a dice game

Consider the simplest possible case where $n = m = 2$. And we will assume that Alice has chosen to just use the first die. So we can ignore her second die, as it is not used. This allows us to easily see all the possible cases:

$$\begin{array}{cccc}A_1&B_1&B_2&A\text{ Score}&B\text{ Score}&\text{Winner}\\\hline 1&1&1&2&2&-\\ 1&1&2&2&3&B\\ 1&2&1&2&3&B\\ 1&2&2&2&4&B\\ 2&1&1&4&2&A\\ 2&1&2&4&3&A\\ 2&2&1&4&3&A\\ 2&2&2&4&4&-\end{array}$$ Since all these are equally likely, Alice has a 50% chance of winning. To get a better feel, you might want to do the similar calculation $n = m = 3$, which has cases where Alice's outcome is not entirely decided (barring ties) by her own roll.

But what you will find is a symmetry: for every case where Alice wins because she rolled high, there is a balancing case where Bob wins because she rolled low, both equally likely. So no, this does not provide Alice with any advantage, but neither does it supply Bob with one.